NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXI11:THE P-BLOCK ELEMENTS
317031
\({{\text{C}}_{{\text{12}}}}{{\text{H}}_{{\text{22}}}}{{\text{O}}_{{\text{11}}}}\xrightarrow{{{\text{conc}}{\text{. }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}}}\) Products. Product(s) of the reaction is/are
317032
On addition of conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to a chloride salt, colourless fumes are evolved but in case of iodide salt, violet fumes come out. This is because:
1 \(\mathrm{H}_{2} \mathrm{SO}_{4}\) reduces \(\mathrm{HI}\) to \(\mathrm{I}_{2}\)
2 \(\mathrm{HI}\) is of violet colour
3 HI gets oxidised to \(\mathrm{I}_{2}\)
4 HI changes to \(\mathrm{HIO}_{3}\)
Explanation:
Hydrogen iodide (HI) is more stronger oxidising agent than \(\mathrm{H}_{2} \mathrm{SO}_{4}\). So, it reduces \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to \(\mathrm{SO}_{2}\) and itself oxidises to \(\mathrm{I}_{2}\). \(\mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{HI} \rightarrow \mathrm{SO}_{2}+\underset{\substack{\text { Violet } \\ \text { colour }}}{\mathrm{I}_{2}}+2 \mathrm{H}_{2} \mathrm{O}\)
CHXI11:THE P-BLOCK ELEMENTS
317033
\(\mathrm{H}_{2} \mathrm{SO}_{4}\) reacts with sugar and acts as
1 A dehydrating agent
2 An oxidising agent
3 As sulphonating agent
4 None of these
Explanation:
Conceptual Questions
CHXI11:THE P-BLOCK ELEMENTS
317034
The number of electrons present in the species used as catalyst in lead chamber process to manufacture \({\mathrm{\mathrm{H}_{2} \mathrm{SO}_{4}}}\) is ____ .
1 14
2 15
3 16
4 18
Explanation:
NO acts as a catalyst in lead chamber process. It contains 15 electrons ( 7 from \({\mathrm{\mathrm{N}, 8}}\) from O ).
317031
\({{\text{C}}_{{\text{12}}}}{{\text{H}}_{{\text{22}}}}{{\text{O}}_{{\text{11}}}}\xrightarrow{{{\text{conc}}{\text{. }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}}}\) Products. Product(s) of the reaction is/are
317032
On addition of conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to a chloride salt, colourless fumes are evolved but in case of iodide salt, violet fumes come out. This is because:
1 \(\mathrm{H}_{2} \mathrm{SO}_{4}\) reduces \(\mathrm{HI}\) to \(\mathrm{I}_{2}\)
2 \(\mathrm{HI}\) is of violet colour
3 HI gets oxidised to \(\mathrm{I}_{2}\)
4 HI changes to \(\mathrm{HIO}_{3}\)
Explanation:
Hydrogen iodide (HI) is more stronger oxidising agent than \(\mathrm{H}_{2} \mathrm{SO}_{4}\). So, it reduces \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to \(\mathrm{SO}_{2}\) and itself oxidises to \(\mathrm{I}_{2}\). \(\mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{HI} \rightarrow \mathrm{SO}_{2}+\underset{\substack{\text { Violet } \\ \text { colour }}}{\mathrm{I}_{2}}+2 \mathrm{H}_{2} \mathrm{O}\)
CHXI11:THE P-BLOCK ELEMENTS
317033
\(\mathrm{H}_{2} \mathrm{SO}_{4}\) reacts with sugar and acts as
1 A dehydrating agent
2 An oxidising agent
3 As sulphonating agent
4 None of these
Explanation:
Conceptual Questions
CHXI11:THE P-BLOCK ELEMENTS
317034
The number of electrons present in the species used as catalyst in lead chamber process to manufacture \({\mathrm{\mathrm{H}_{2} \mathrm{SO}_{4}}}\) is ____ .
1 14
2 15
3 16
4 18
Explanation:
NO acts as a catalyst in lead chamber process. It contains 15 electrons ( 7 from \({\mathrm{\mathrm{N}, 8}}\) from O ).
317031
\({{\text{C}}_{{\text{12}}}}{{\text{H}}_{{\text{22}}}}{{\text{O}}_{{\text{11}}}}\xrightarrow{{{\text{conc}}{\text{. }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}}}\) Products. Product(s) of the reaction is/are
317032
On addition of conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to a chloride salt, colourless fumes are evolved but in case of iodide salt, violet fumes come out. This is because:
1 \(\mathrm{H}_{2} \mathrm{SO}_{4}\) reduces \(\mathrm{HI}\) to \(\mathrm{I}_{2}\)
2 \(\mathrm{HI}\) is of violet colour
3 HI gets oxidised to \(\mathrm{I}_{2}\)
4 HI changes to \(\mathrm{HIO}_{3}\)
Explanation:
Hydrogen iodide (HI) is more stronger oxidising agent than \(\mathrm{H}_{2} \mathrm{SO}_{4}\). So, it reduces \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to \(\mathrm{SO}_{2}\) and itself oxidises to \(\mathrm{I}_{2}\). \(\mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{HI} \rightarrow \mathrm{SO}_{2}+\underset{\substack{\text { Violet } \\ \text { colour }}}{\mathrm{I}_{2}}+2 \mathrm{H}_{2} \mathrm{O}\)
CHXI11:THE P-BLOCK ELEMENTS
317033
\(\mathrm{H}_{2} \mathrm{SO}_{4}\) reacts with sugar and acts as
1 A dehydrating agent
2 An oxidising agent
3 As sulphonating agent
4 None of these
Explanation:
Conceptual Questions
CHXI11:THE P-BLOCK ELEMENTS
317034
The number of electrons present in the species used as catalyst in lead chamber process to manufacture \({\mathrm{\mathrm{H}_{2} \mathrm{SO}_{4}}}\) is ____ .
1 14
2 15
3 16
4 18
Explanation:
NO acts as a catalyst in lead chamber process. It contains 15 electrons ( 7 from \({\mathrm{\mathrm{N}, 8}}\) from O ).
317031
\({{\text{C}}_{{\text{12}}}}{{\text{H}}_{{\text{22}}}}{{\text{O}}_{{\text{11}}}}\xrightarrow{{{\text{conc}}{\text{. }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}}}\) Products. Product(s) of the reaction is/are
317032
On addition of conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to a chloride salt, colourless fumes are evolved but in case of iodide salt, violet fumes come out. This is because:
1 \(\mathrm{H}_{2} \mathrm{SO}_{4}\) reduces \(\mathrm{HI}\) to \(\mathrm{I}_{2}\)
2 \(\mathrm{HI}\) is of violet colour
3 HI gets oxidised to \(\mathrm{I}_{2}\)
4 HI changes to \(\mathrm{HIO}_{3}\)
Explanation:
Hydrogen iodide (HI) is more stronger oxidising agent than \(\mathrm{H}_{2} \mathrm{SO}_{4}\). So, it reduces \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to \(\mathrm{SO}_{2}\) and itself oxidises to \(\mathrm{I}_{2}\). \(\mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{HI} \rightarrow \mathrm{SO}_{2}+\underset{\substack{\text { Violet } \\ \text { colour }}}{\mathrm{I}_{2}}+2 \mathrm{H}_{2} \mathrm{O}\)
CHXI11:THE P-BLOCK ELEMENTS
317033
\(\mathrm{H}_{2} \mathrm{SO}_{4}\) reacts with sugar and acts as
1 A dehydrating agent
2 An oxidising agent
3 As sulphonating agent
4 None of these
Explanation:
Conceptual Questions
CHXI11:THE P-BLOCK ELEMENTS
317034
The number of electrons present in the species used as catalyst in lead chamber process to manufacture \({\mathrm{\mathrm{H}_{2} \mathrm{SO}_{4}}}\) is ____ .
1 14
2 15
3 16
4 18
Explanation:
NO acts as a catalyst in lead chamber process. It contains 15 electrons ( 7 from \({\mathrm{\mathrm{N}, 8}}\) from O ).
